Added the Kaplansky Density Theorem.

src/fa/lc/beqtensor.tex

@@ -58,5 +58,44 @@
     is \textit{the} \textbf{injective cross norm} on $E \otimes_\eps F$. 
 \end{definition}
 
+\begin{definition}[Integral Bilinear Form]
+\label{definition:integral-bilinear-form}
+    Let $E, F$ be locally convex spaces over $K \in \RC$ and $\lambda \in L^2(E, F; K)$ be a bilinear form, then $\lambda$ is \textbf{integral} if there exists equicontinuous subsets $S \subset E^*$ and $T \subset F^*$, and a Radon measure $\mu \in M_R(S \times T; K)$ such that
+    \[
+    \lambda(x, y) = \int_{S \times T} \dpn{x, \phi}{E} \dpn{y, \psi}{F} \mu(d\phi, d\psi)
+    \]
+
+    for all $x, y \in E$. 
+
+    The set $I(E, F)$ is the \textbf{space of integral bilinear forms} on $E$ and $F$. 
+\end{definition}
+
+
+\begin{theorem}
+\label{theorem:injective-dual-bilinear-form}
+    Let $E, F$ be locally convex spaces over $K \in \RC$, then $(E \wh \otimes_\eps F)^* = I(E, F)$. 
+\end{theorem}
+\begin{proof}
+    Let $\lambda \in (E \wh \otimes_\eps F)^*$, then there exists equicontinuous subsets $S \subset E^*$ and $T \subset F^*$ such that for each $x \in E$ and $y \in F$, 
+    \[
+    |\lambda(x, y)| \le \sup_{\phi \in S}\sup_{\psi \in T} |\dpn{x, \phi}{E} \dpn{y, \psi}{F}|
+    \]
+
+    For any $(x, y) \in E \times F$ and $(\phi, \psi) \in S \times T$, let $f_{xy}(\phi, \psi) = \dpn{x, \phi}{E} \dpn{y, \psi}{F}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Lambda \in C(S \times T; K)^*$ such that the following diagram commutes:
+    \[
+    \xymatrix{
+    & C(S \times T; K) \ar@{->}[rd]^{\Lambda} &  \\
+    E \times F \ar@{->}[ru]^{{(x, y) \mapsto f_{xy}}} \ar@{->}[rr]_{\lambda} &  & K
+    }
+    \]
+
+    Now, since $S$ and $T$ are equicontinuous, using the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, assume without loss of generality that $S$ and $T$ are weak*-compact. In which case, by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(S \times T; K)$ such that $\Lambda(f) = \int_{S \times T} f d\mu$ for all $f \in C(S \times T; K)$. Therefore
+    \[
+    \lambda(x, y) = \Lambda(f_{xy}) = \int_{S \times T}f_{xy} d\mu = \int_{S \times T} \dpn{x, \phi}{E}\dpn{y, \psi}{F} d\mu
+    \]
+
+    for all $(x, y) \in E \times F$, and $\lambda \in I(E; F)$. 
+\end{proof}
+
 
 

src/fa/norm/separable.tex

@@ -12,22 +12,8 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\seq{x_n} \subset E$ be a dense subset. For each $N \in \natp$, let
+    (1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $S$ is embedded as a subspace of $\real^D$. By \autoref{theorem:uniform-metrisable}, $\real^D$ is metrisable. By \autoref{proposition:separable-product}, $\real^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}. 
-    \[
+    
-    T_N: S \to \real^N \quad y \mapsto (\dpn{x_1, y}{\lambda}, \cdots, \dpn{x_N, y}{\lambda})
-    \]
-
-    Since $\real^N$ is separable, $T_N(S)$ is separable by \autoref{proposition:separable-metric-space}. Thus there exists $\bracs{y_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_Ny_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$.
-
-    Let $y \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, 
-    \[
-    |\dpn{x_n, y_{N, k_N}}{\lambda} - \dpn{x_n, y}{\lambda}| \le \frac{1}{N}
-    \] 
-
-    Thus for each $N \in \natp$, $\dpn{x_n, y_{N, k_N}}{\lambda} \to \dpn{x_n, y}{\lambda}$ as $N \to \infty$. Since $y_{N, k_N} \to y$ pointwise on a dense subset of $E$ and $\bracsn{y_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $y_{N, k_N} \to y$ in the $\sigma(F, E)$-topology by \autoref{proposition:strong-operator-dense}. 
-
-    (2): Let $\seq{x_n} \subset E$ be a dense subset, then by \autoref{proposition:strong-operator-dense}, the $\sigma(F, E)$-topology on $S$ is induced by $\seq{x_n}$, and hence metrisable by \autoref{theorem:uniform-metrisable}. 
-
     (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. 
 \end{proof}
 

src/fa/notation.tex

@@ -29,6 +29,7 @@
     $E \,\wh{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $p \otimes q$ & Cross seminorm of $p$ and $q$. & \autoref{definition:cross-seminorm} \\
     $N(E; F)$ & Nuclear mappings from $E$ to $F$. & \autoref{definition:nuclear-operator-normed} \\ 
+    $I(E, F)$ & Integral bilinear forms on $E \times F$. & \autoref{definition:integral-bilinear-form} \\ 
     % ---- Order Structures ----
     $x \vee y$, $x \wedge y$ & $\sup$ and $\inf$ in vector lattice. & \autoref{definition:vector-lattice} \\
     $|x|$ & Absolute value $x \vee (-x)$ in a vector lattice. & \autoref{definition:order-absolute-value} \\

src/fa/tvs/bilinear.tex

@@ -94,13 +94,13 @@
 
     Now, let $S \in \sigma$, then $\bracsn{\lambda(x, \cdot)|x \in S}$ is equicontinuous by the $\sigma$-hypocontinuity of $\lambda$. For any $U \in \cn_0(G)$, there exists $V \in \cn_0(F)$ such that $\bigcup_{x \in S}\lambda(x, V \cap F_0) \subset U$. By \autoref{proposition:closure-of-image}, $\bigcup_{x \in S}\Lambda(x, \ol V) \subset \ol U$. Thus \autoref{proposition:tvs-good-neighbourhood-base} implies that $\bracsn{\Lambda(x, \cdot)|x \in S}$ is equicontinuous as well. 
     
-    For each $x_0 \in \ol S$ and $y_0 \in F$, there exists $T \in \tau$ with $y_0 \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_0, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, 
+    For each $x_0 \in \ol S$ and $y_0 \in F$, there exists $T \in \tau$ with $y_0 \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_0, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. Thus
-    \begin{enumerate}
+    \[
-        \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}$. 
+    \bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}
-        \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous. 
+    \]
-    \end{enumerate}
+    
-
+    
-    so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. 
+    By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous, so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. 
 \end{proof}
 
 

src/op/notation.tex

@@ -19,6 +19,7 @@
     $P(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
     $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
     $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
+    $U(T)$ & Cayley transform of $T$. & \autoref{definition:cayley-transform-bounded} \\ 
 
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\

src/op/vn/cayley.tex

@@ -0,0 +1,53 @@
+\section{The Cayley Transform}
+\label{section:cayley-transform}
+
+\begin{definition}[Cayley Transform]
+\label{definition:cayley-transform-bounded}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $-i \not\in \sigma_A(H)$, then $U(T) = (T - i)(T + i)^{-1}$ is the \textbf{Cayley transform} of $T$. 
+\end{definition}
+
+\begin{theorem}
+\label{theorem:cayley-sa-uni}
+    Let $H$ be a complex Hilbert space and $U_1$ be the set of unitary operators on $H$ with $1$ not in their spectrum, then the Cayley transform $T \mapsto (T - i)(T + i)^{-1}$ is a strong-operator continuous bijection between $B(H)_{sa}$ and $U_1$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 19.3]{Zhu}}}. ]
+    Let $T \in B(H)$ be self-adjoint. By \autoref{proposition:self-adjoint-spectrum}, $\sigma_{B(H)}(T) \subset \real$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, 
+    \[
+    \sigma_{B(H)}[(T-i)(T+i)^{-1}] \subset \bracsn{(t - i)/(t + i)|t \in \real} \subset \partial B_\complex(0, 1) \setminus \bracsn{1}
+    \]
+    
+    Hence $(T - i)(T+i)^{-1}$ is a well-defined unitary element of $B(H)$ whose spectrum does not contain $1$. 
+
+    Since the mapping $t \mapsto -i(t + 1)/(t - 1)$ is the inverse of $t \mapsto (t - i)/(t + i)$ on $\partial B_\complex(0, 1)$, the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous} implies that $T \mapsto -i(T + I)(T - I)^{-1}$ is the inverse of the Cayley transform on $U_1$. 
+
+    For any self-adjoint elements $S, T \in B(H)$,
+    \begin{align*}
+        U(S) - U(T) &= (S + i)^{-1}(S - i) - (T - i)(T+i)^{-1} \\
+        &= (S + i)^{-1}[(S - i)(T + i)  - (S + i)(T - i)](T + i)^{-1} \\
+        &= 2i(S + i)^{-1}(S - T)(T + i)^{-1}
+    \end{align*}
+
+    so for any $x \in H$, 
+    \[
+    \normn{[U(S) - U(T)]x}_H \le 2\normn{(S + i)^{-1}}_{B(H)} \cdot \normn{(S - T)(T+i)^{-1}x}_H
+    \]
+
+    By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\normn{(S+i)^{-1}}_{B(H)} \le 1$. Therefore the Cayley transform is strong-operator continuous. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:functional-calculus-c0-self-adjoint}
+    Let $H$ be a complex Hilbert space and $f \in C_0(\real; \complex)$, then the mapping $T \mapsto f(T)$ is strong-operator continuous on $B(H)_{sa}$. 
+\end{corollary}
+\begin{proof}
+    Let
+    \[
+    g: \partial B_\complex(0, 1) \to \complex \quad z \mapsto \begin{cases}
+        f(-i(z+1)/(z-1)) &z \ne 1 \\
+        0 &z = 1
+    \end{cases}
+    \]
+
+    then since $f \in C_0(\real; \complex)$, $g \in C(\partial B_\complex(0, 1); \complex)$. For each $T \in B(H)_{sa}$, $f(T) = g(U(T))$. By \autoref{proposition:bh-adjoint-strong-continuous}, the mapping $U \mapsto g(U)$ is strong-operator continuous on the set of unitary operators on $H$. By \autoref{theorem:cayley-sa-uni}, $T \mapsto f(T)$ is the composition of two strong-operator continuous mappings. 
+\end{proof}
+

src/op/vn/index.tex

@@ -2,4 +2,5 @@
 \label{chap:von-neumann-algebras}
 
 \input{./topologies.tex}
+\input{./cayley.tex}
 \input{./vn.tex}

src/op/vn/topologies.tex

@@ -16,6 +16,19 @@ Depending on the topology placed on $H \otimes H$, and the corresponding complet
     By \autoref{proposition:projective-tensor-product-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}. 
 \end{proof}
 
+A natural topology consistent with the ultraweak topology would be the ultrastrong topology. 
+
+\begin{definition}[Ultrastrong Topology]
+\label{definition:bh-ultrastrong-topology}
+    Let $H$ be a complex Hilbert space. For each $x = \seq{x_n} \in L^2(\natp; H)$, let
+    \[
+    \Phi_x: B(H) \to L^2(\natp; H) \quad (\Phi_xT)_n = Tx_n
+    \]
+
+    then the \textbf{ultrastrong}/\textbf{$\sigma$-strong} topology on $B(H)$ is the topology generated by the maps $\bracsn{\Phi_x|x \in L^2(\natp; H)}$. 
+\end{definition}
+
+
 Seeing that $B(H)$ is a dual Banach space, the following fact is immediate:
 
 \begin{proposition}
@@ -36,7 +49,6 @@ Now, a few facts about the more familiar operator topologies:
         \item Every bounded subset of $B(H)$ is relatively compact in the weak operator topology. 
         \item The composition map $(S, T) \mapsto ST$ is separately continuous in the strong and weak operator topologies. 
         \item The composition map $(S, T) \mapsto ST$ is left-hypocontinuous with respect to the strong operator topology and strong-operator bounded subsets of $B(H)$. 
-        \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -45,5 +57,55 @@ Now, a few facts about the more familiar operator topologies:
     (4): By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, every strong-operator bounded subset of $B(H)$ is equicontinuous. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:bh-adjoint-strong-continuous}
+    Let $H$ be a complex Hilbert space, then:
+    \begin{enumerate}
+        \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
+        \item $T \mapsto T^*$ restricted to the normal operators is continuous in the strong operator topology. 
+        \item For any $f \in C(\complex; \complex)$, the mapping $T \mapsto f(T)$ restricted to any bounded set of normal operators is continuous in the strong operator topology. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Section 19.1]{Zhu}}}. ]
+    (2): Let $S, T \in B(H)$, then
+    \begin{align*}
+        \normn{(S^* - T^*)x}_H^2 &= \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, ST^*x}{H} - \dpn{ST^*x, x}{H} \\
+        &\le \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, TT^*x}{H} - \dpn{TT^*x, x}{H} \\
+        &+ |\dpn{x, (T - S)T^*x}{H}| + |\dpn{(T - S)T^*x, x}{H}| \\
+        &\le |\normn{S^*x}_H^2 - \normn{T^*x}_H^2| + 2\norm{x}_H\normn{(T - S)T^*x}_H
+    \end{align*}
 
+    Now, if $S$ and $T$ are normal, then $\normn{S^*x}_H = \norm{Sx}_H$ and $\norm{T^*x}_H = \norm{Tx}_H$, so
+    \begin{align*}
+    |\norm{S^*x}_H^2 - \norm{T^*x}_H^2| &= |\norm{Sx}_H^2 - \norm{Tx}_H^2| \\
+    &\le \norm{(S - T)x}_H (\norm{Sx}_H + \norm{Tx}_H) \\
+    &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H)
+    \end{align*}
+
+    Therefore
+    \begin{align*}
+    \normn{(S^* - T^*)x}_H^2 &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H) \\
+    &+ 2\norm{x}_H\normn{(T - S)T^*x}_H
+    \end{align*}
+    
+    and the adjoint map restricted to normal operators is continuous in the strong operator topology. 
+
+    (3): Let $S, T \in B_{B(H)}(0, 1)$ and $x \in H$ and $n \in \natp$, then
+    \begin{align*}
+    \normn{(S^n - T^n)x}_H &\le \sum_{k = 0}^{n-1}\normn{S^{n-1-k}(S - T)T^kx}_{H} \\
+    &\le \sum_{k = 0}^{n - 1}\normn{(S - T)T^kx}_H
+    \end{align*}
+    
+
+    so the mapping $T \mapsto T^n$ on $B_{B(H)}(0, 1)$ is continuous in the strong operator topology. By (2), the mapping $T \mapsto p(T, T^*)$ is strong-operator continuous for all $p \in \complex[z, \ol z]$. 
+
+    By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exist polynomials $p_n \in \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\ol{B_\complex(0, 1)}$. For any $T \in B_{B(H)}(0, 1)$, $x \in H$, and $n \in \natp$, 
+    \begin{align*}
+    \norm{[f(T) - p_n(T)]x}_H &\le \norm{f(T) - p_n(T)}_{B(H)} \cdot \norm{x}_H \\
+    &\le \norm{x}_H  \cdot \sup_{z \in \ol{B_\complex(0, 1)}}|f(z) - p_n(z)|
+    \end{align*}
+    
+
+    by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}. Thus $f$ is a uniform limit of strong-operator continuous functions on $B_{B(H)}(0, 1)$, and as such also strong-operator continuous by \autoref{proposition:uniform-limit-continuous}. 
+\end{proof}
 

src/op/vn/vn.tex

@@ -141,6 +141,20 @@
     Now, since $T \in A''$, $TP = PT$ as well, so $M$ is a reducing subspace for $T$. As $A$ is unital, $x \in M$, so $Tx \in M = \ol{\bracsn{Sx|S \in A}}$. 
 \end{proof}
 
+\begin{lemma}[Amplification]
+\label{lemma:bh-amplification}
+    Let $H$ be a complex Hilbert space, $n \in \natp$, and
+    \[
+    \pi: B(H) \to B(H^n) \quad [\pi(T)(x)]_n = Tx_n
+    \]
+
+    then for each $\seqf{x_j} \subset H$, 
+    \[
+    \max_{1 \le j \le n}\norm{Tx_j}_H \le \norm{\pi(T)(x)}_{H^n} \le n \max_{1 \le j \le n}\norm{Tx_j}_H
+    \]
+\end{lemma}
+
+
 \begin{theorem}[Von Neumann's Bicommutant Theorem]
 \label{theorem:bicommutant}
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, then $A''$ is the strong-operator closure of $A$. 
@@ -162,3 +176,76 @@
     so $T \in \ol{A}^{\text{\small SOT}}$. 
 \end{proof}
 
+\begin{definition}[Von Neumann Algebra]
+\label{definition:von-neumann-algebra}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a \textbf{von Neumann algebra acting on $H$} if $A$ is closed in the strong operator topology. 
+\end{definition}
+
+\begin{theorem}[Kaplansky Density Theorem]
+\label{theorem:kaplansky-density}
+    Let $H$ be a Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra, and $B$ be the strong-operator closure of $A$, then:
+    \begin{enumerate}
+        \item $\ol{B_{A_{sa}}(0, 1)}$ is strong-operator dense in $\ol{B_{B_{sa}}(0, 1)}$.
+        \item $\bracsn{T \in \ol{B_{A}(0, 1)}|T \ge 0}$ is strong-operator dense in $\bracsn{T \in \ol{B_{B}(0, 1)}|T \ge 0}$.
+        \item $\ol{B_{A}(0, 1)}$ is strong-operator dense in $\ol{B_{B}(0, 1)}$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 19.5]{Zhu}}}. ]
+    (1): Let $T \in \ol{B_{B_{sa}}(0, 1)}$ and $\angles{T_\gamma}_{\gamma \in C} \subset A$ be a net such that $T_\gamma \to T$ in the weak operator topology. For each $\gamma \in C$, let $T_\gamma' = (T_\gamma + T_\gamma^*)/2$, then $T_\gamma' \to T$ in the weak operator topology by continuity of the adjoint map in the weak operator topology. 
+
+    As $A_{sa}$ is a subspace of $B(H)$, its strong and weak-operator closures coincide. Thus there exists a net $\angles{S_\gamma}_{\gamma \in C} \subset A_{sa}$ such that $S_\gamma \to T$ in the strong operator topology. In which case, let
+    \[
+    f: \real \to \real \quad t \mapsto \begin{cases}
+        t &t \in [-1, 1] \\
+        1/t &t \in \real \setminus [-1, 1]
+    \end{cases}
+    \]
+
+    then $f \in C_0(\real; \real)$. By \autoref{corollary:functional-calculus-c0-self-adjoint}, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)} \le 1$, $\sigma_{B}(T) \subset [-1, 1]$. Thus $f(T) = T$, and $f(S_\gamma) \to T$ in the strong operator topology. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $f(S_\gamma)$ is in the closed unit ball of $A_{sa}$ for all $\gamma \in C$. Therefore the closed unit ball of $A_{sa}$ is strong-operator dense in the closed unit ball of $B_{sa}$. 
+
+    (2): Let $T \in \ol{B_{B}(0, 1)}$ with $T \ge 0$ and $\angles{T_\gamma}_{\gamma \in C} \subset A_{sa}$ be a net such that $T_\gamma \to T$ in the strong operator topology. Define
+    \[
+    f: \real \to \real \quad t \mapsto \begin{cases}
+        0 &t \le 0 \\
+        t &t \in [0, 1] \\
+        1/t &t \ge 1
+    \end{cases}
+    \]
+
+    then $f \in C_0(\real; [0, \infty))$. Since $f \ge 0$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} then implies that $\norm{f(T_\gamma)}_{B(H)} \le 1$ and $f(T_\gamma) \ge 0$ for all $\gamma \in C$. By \autoref{corollary:functional-calculus-c0-self-adjoint}, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)} \le 1$ and $T \ge 0$, $\sigma_{B}(T) \subset [0, 1]$ by \autoref{proposition:positive-spectrum}. Thus $f(T) = T$, and $f(T_\gamma) \to T$ in the strong operator topology. 
+
+    (3): For each $\mathcal{T} \subset B(H)$, let
+    \[
+    M_2(\mathcal{T}) = \bracs{\begin{bmatrix} Q & R \\ S & T \end{bmatrix} \bigg | Q, R, S, T \in \mathcal{T}}
+    \]
+
+    then $M_2(B)$ is the strong-operator closure of $M_2(A)$ in $B(H^2)$. For each $T \in \ol{B_{B}(0, 1)}$, let
+    \[
+    T' = \begin{bmatrix} 0 & T \\ T^* & 0 \end{bmatrix}
+    \]
+
+    then $T' \in \ol{B_{M_2(B)_{sa}}(0, 1)}$. By (1), there exists a net $\angles{(R_\gamma, S_\gamma, T_\gamma)}_{\gamma \in C} \subset A^3$ such that:
+    \begin{enumerate}
+        \item For each $\gamma \in C$, 
+        \[
+        \norm{\begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix}}_{B(H^2)} \le 1
+        \]
+
+        In particular, $\norm{T_\gamma}_{B(H)} \le 1$.
+        \item With respect to the strong operator topology on $B(H^2)$,
+        \[
+        \begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix} \to T'
+        \]
+    \end{enumerate}
+
+    Therefore $\angles{T_\gamma} \subset \ol{B_A(0, 1)}$ is a net that converges to $T$ in the strong-operator topology. 
+    
+\end{proof}
+
+
+\begin{remark}
+\label{remark:kaplansky-unitary}
+    The Kaplansky Density Theorem should also apply to the unitary case. Unfortunately, it seems like that the Borel functional calculus is required for an easier proof, so it will be postponed for now. 
+\end{remark}
+
+